\newproblem{lay:3_1_42}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 3.1.42}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Let $\mathbf{u}=\begin{pmatrix}a\\b\end{pmatrix}$ and $\mathbf{v}=\begin{pmatrix}c\\0\end{pmatrix}$, where $a,b,c$ are positive (for simplicity).
	Compute the area of the parallelogram determined by $\mathbf{u}$, $\mathbf{v}$, $\mathbf{u}+\mathbf{v}$ and $\mathbf{0}$, and compute the determinants
	of the matrices $\begin{pmatrix}\mathbf{u} & \mathbf{v} \end{pmatrix}$ and $\begin{pmatrix}\mathbf{v} & \mathbf{u} \end{pmatrix}$. Draw a picture and explain
	what you find.
}{
   % Solution
	The area of the parallelogram is base times height. In this case:
	\begin{center}
		\includegraphics[scale=0.5]{Tema4/lay_3_1_42.eps}\\
		$A=cb$
	\end{center}
	The determinant of $\begin{pmatrix}\mathbf{u} & \mathbf{v} \end{pmatrix}$ is
	\begin{center}
		$\left|\begin{array}{cc} a & c \\ b & 0\end{array}\right|=a\cdot 0 - bc=-bc$
	\end{center}
	The determinant of $\begin{pmatrix}\mathbf{v} & \mathbf{u} \end{pmatrix}$ is
	\begin{center}
		$\left|\begin{array}{cc} c & a \\ 0 & b\end{array}\right|=cb-a\cdot 0=cb$
	\end{center}
	We see that $A=\mathrm{abs}\left(\left|\begin{pmatrix}\mathbf{u} & \mathbf{v} \end{pmatrix}\right|\right)=
	   \mathrm{abs}\left(\left|\begin{pmatrix}\mathbf{v} & \mathbf{u} \end{pmatrix}\right|\right)$
}
\useproblem{lay:3_1_42}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
